Selina Solutions Concise Maths Class 7 Chapter 14 Lines and Angles (Including Construction of Angles) Exercise 14B contains 100% accurate answers, strictly based on the ICSE syllabus. The concept of a transversal, parallel lines and the conditions of parallelism are the topics which are covered under this exercise. The important shortcut tricks and formulas are explained in simple language to impart better understanding among students. Selina Solutions Concise Maths Class 7 Chapter 14 Lines and Angles (Including Construction of Angles) Exercise 14B, PDF can be downloaded from the links available below.

## Selina Solutions Concise Maths Class 7 Chapter 14: Lines and Angles (Including Construction of Angles) Exercise 14B Download PDF

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#### Exercise 14B page: 166

**1. In questions 1 and 2, given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles:**

**(i) âˆ 3 and âˆ 6
(ii) âˆ 2 and âˆ 4
(iii) âˆ 3 and âˆ 7
(iv) âˆ 2 and âˆ 7
(v) âˆ 4 andâˆ 6
(vi) âˆ 1 and âˆ 8
(vii) âˆ 1 and âˆ 5
(viii) âˆ 1 and âˆ 4
(ix) âˆ 5 and âˆ 7**

**Solution:**

(i) âˆ 3 and âˆ 6 are interior alternate angles.

(ii) âˆ 2 and âˆ 4 are adjacent angles.

(iii) âˆ 3 and âˆ 7 are corresponding angles.

(iv) âˆ 2 and âˆ 7 are exterior alternate angles.

(v) âˆ 4 andâˆ 6 are allied or co-interior angles.

(vi) âˆ 1 and âˆ 8 are exterior alternate angles.

(vii) âˆ 1 and âˆ 5 are corresponding angles.

(viii) âˆ 1 and âˆ 4 are vertically opposite angles.

(ix) âˆ 5 and âˆ 7 are adjacent angles.

**2. (i) âˆ 1 and âˆ 4
(ii) âˆ 4 and âˆ 7
(iii) âˆ 10 and âˆ 12
(iv) âˆ 7 and âˆ 13
(v) âˆ 6 and âˆ 8
(vi) âˆ 11 and âˆ 8
(vii) âˆ 7 and âˆ 9
(viii) âˆ 4 and âˆ 5
(ix) âˆ 4 and âˆ 6
(x) âˆ 6 and âˆ 7
(xi) âˆ 2 and âˆ 13**

**Solution:**

(i) âˆ 1 and âˆ 4 are vertically opposite angles.

(ii) âˆ 4 and âˆ 7 are interior alternate angles.

(iii) âˆ 10 and âˆ 12 are vertically opposite angles.

(iv) âˆ 7 and âˆ 13 are corresponding angles.

(v) âˆ 6 and âˆ 8 are vertically opposite angles.

(vi) âˆ 11 and âˆ 8 are allied or co-interior angles.

(vii) âˆ 7 and âˆ 9 are vertically opposite angles.

(viii) âˆ 4 and âˆ 5 are adjacent angles.

(ix) âˆ 4 and âˆ 6 are allied or co-interior angles.

(x) âˆ 6 and âˆ 7 are adjacent angles.

(xi) âˆ 2 and âˆ 13 are allied or co-interior angles.

**3. In the following figures, the arrows indicate parallel lines. State which angles are equal. Give reasons.**

**Solution:**

(i) From the figure (i)

a = b are corresponding angles

b = c are vertically opposite angles

a = c are alternate angles

So we get

a = b = c

(ii) From the figure (ii)

x = y are vertically opposite angles

y = l are alternate angles

x = l are corresponding angles

1 = n are vertically opposite angles

n = r are corresponding angles

So we get

x = y = l = n = r

Similarly

m = k are vertically opposite angles

k = q are corresponding angles

Hence, m = k = q.

**4. In the given figure, find the measure of the unknown angles:**

**Solution:**

From the figure

a = d are vertically opposite angles

d = f are corresponding angles

f = 110^{0} are vertically opposite angles

So we get

a = d = f = 110^{0}

We know that

e + 110^{0} = 180^{0} are linear pair of angles

e = 180 â€“ 110 = 70^{0}

b = c are vertically opposite angles

b = e are corresponding angles

e = g are vertically opposite angles

So we get

b = c = e = g = 70^{0}

Therefore, a = 110^{0}, b = 70^{0}, c = 70^{0}, d = 110^{0}, e = 70^{0}, f = 110^{0} and g = 70^{0}.

**5. Which pair of the dotted line, segments, in the following figures, are parallel. Give reason:**

**Solution:**

(i) From the figure (i)

If the lines are parallel we get 120^{0} + 50^{0} = 180^{0}

There are co-interior angles where 170^{0}Â â‰ 180^{0}

Therefore, they are not parallel lines.

(ii) From the figure (ii)

âˆ 1 = 45^{0} are vertically opposite angles

We know that the lines are parallel if

âˆ 1 + 135^{0} = 180^{0} are co-interior angles

Substituting the values

45^{0} + 135^{0} = 180^{0}

180^{0} = 180^{0} which is true

Therefore, the lines are parallel.

(iii) From the figure (iii)

The lines are parallel if corresponding angles are equal

Here 120^{0}Â â‰ 130^{0}

Hence, lines are not parallel.

(iv) âˆ 1 = 110^{0} are vertically opposite angles

We know that if lines are parallel

âˆ 1 + 70^{0} = 180^{0} are co-interior angles

Substituting the values

110^{0} + 70^{0} = 180^{0}

180^{0} = 180^{0} which is correct

Therefore, the lines are parallel.

(v) âˆ 1 + 100^{0} = 180^{0}

So we get

âˆ 1 = 180^{0} â€“ 100^{0} = 80^{0} which is a linear pair

Here the lines l_{1} and l_{2} are parallel if âˆ 1 = 70^{0}

But here âˆ 1 = 80^{0} which is not equal to 70^{0}

So the lines l_{1} and l_{2} are not parallel

Now, l_{3} and l_{5} will be parallel if corresponding angles are equal

But here, the corresponding angles 80^{0} â‰ 70^{0}

Hence, l_{3} and l_{5} are not parallel.

We know that, in l_{2} and l_{4}

âˆ 1 = 80^{0} are alternate angles

80^{0} = 80^{0} which is true

Hence, l_{2} and l_{4} are parallel.

(vi) Two lines are parallel if alternate angles are equal

Here, 50^{0}Â â‰ 40^{0 }which is not true

Hence, the lines are not parallel.

**6. In the given figures, the directed lines are parallel to each other. Find the unknown angles.**

**Solution:**

(i) If the lines are parallel

a = b are corresponding angles

a = c are vertically opposite angles

a = b = c

Here b = 60^{0} are vertically opposite angles

Therefore, a = b = c = 60^{0}

(ii) If the lines are parallel

x = z are corresponding angles

z + y = 180^{0} is a linear pair

y = 55^{0} are vertically opposite angles

Substituting the values

z + 55^{0} = 180^{0}

z = 180 â€“ 55 = 125^{0}

If x = z we get x = 125^{0}

Therefore, x = 125^{0}, y = 55^{0} and z = 125^{0}.

(iii) If the lines are parallel

c = 120^{0} (corresponding angles)

a + 120^{0} = 180^{0} are co-interior angles

a = 180 â€“ 120 = 60^{0}

We know that a = b are vertically opposite angles

So b = 60^{0}

Therefore, a = b = 60^{0} and c = 120^{0}.

(iv) If the lines are parallel

x = 50^{0} are alternate angles

y + 120^{0} = 180^{0} are co-interior angles

y = 180 â€“ 120 = 60^{0}

We know that

x + y + z = 360^{0} are angles at a point

Substituting the values

50 + 60 + z = 360

By further calculation

110 + z = 360

z = 360 â€“ 110 = 250^{0}

Therefore, x = 50^{0}, y = 60^{0} and z = 250^{0}.

(v) If the lines are parallel

x + 90^{0} = 180^{0} are co-interior angles

x = 180^{0} â€“ 90^{0} = 90^{0}

âˆ 2 = x

âˆ 2 = 90^{0}

We know that the sum of angles of a triangle

âˆ 1 + âˆ 2 + 30^{0} = 180^{0}

Substituting the values

âˆ 1 + 90^{0} + 30^{0} = 180^{0}

By further calculation

âˆ 1 + 120^{0} = 180^{0}

âˆ 1 = 180 â€“ 120 = 60^{0}

Here âˆ 1 = k are vertically opposite angles

k = 60^{0}

Here âˆ 1 = z are corresponding angles

z = 60^{0}

Here k + y = 180^{0} are co-interior angles

Substituting the values

60^{0} + y = 180^{0}

y = 180 â€“ 60 = 120^{0}

Therefore, x = 90^{0}, y = 120^{0}, z = 60^{0}, k = 60^{0}.

**7. Find x, y and p is the given figures:**

**Solution:**

(i) From the figure (i)

The lines are parallel

x = z are corresponding angles

y = 40^{0} are corresponding angles

We know that

x + 40^{0} + 270^{0} = 360^{0} are the angles at a point

So we get

x + 310^{0} = 360^{0}

x = 360 â€“ 310 = 50^{0}

So z = x = 50^{0}

Here p + z = 180^{0} is a linear pair

By substituting the values

p + 50^{0} = 180^{0}

p = 180 â€“ 50 = 130^{0}

Therefore, x = 50^{0}, y = 40^{0}, z = 50^{0} and p = 130^{0}.

(ii) From the figure (ii)

The lines are parallel

y = 110^{0}Â are corresponding angles

We know that

25^{0} + p + 110^{0} = 180^{0} are angles on a line

p + 135^{0} = 180^{0}

p = 180 â€“ 135 = 45^{0}

We know that the sum of angles of a triangle

x + y + 25^{0} = 180^{0}

x + 110^{0} + 25^{0} = 180^{0}

By further calculation

x + 135^{0} = 180^{0}

x = 180 â€“ 135 = 45^{0}

Therefore, x = 45^{0}, y = 110^{0} and p = 45^{0}.

**8. Find x in the following cases:**

**Solution:**

(i) From the figure (i)

The lines are parallel

2x + x = 180^{0} are co-interior angles

3x = 180^{0}

x = 180/3 = 60^{0}

(ii) From the figure (ii)

The lines are parallel

4x + âˆ 1 = 180^{0} are co-interior angles

âˆ 1 = 5x are vertically opposite angles

Substituting the values

4x + 5x = 180^{0}

So we get

9x = 180^{0}

x = 180/9 = 20^{0}

(iii) From the figure (iii)

The lines are parallel

âˆ 1 + 4x = 180^{0} are co-interior angles

âˆ 1 = x are vertically opposite angles

Substituting the values

x + 4x = 180^{0}

5x = 180^{0}

So we get

x = 180/5 = 36^{0}

(iv) From the figure (iv)

The lines are parallel

2x + 5 + 3x + 55 = 180^{0} are co-interior angles

5x + 60^{0} = 180^{0}

By further calculation

5x = 180 â€“ 60 = 120^{0}

So we get

x = 120/5 = 24^{0}

(v) From the figure (v)

The lines are parallel

âˆ 1 = 2x + 20^{0} are alternate angles

âˆ 1 + 3x + 25^{0} = 180^{0} is a linear pair

Substituting the values

2x + 20^{0} + 3x + 25^{0} = 180^{0}

5x + 45^{0} = 180^{0}

So we get

5x = 180 â€“ 45 = 135^{0}

x = 135/5 = 27^{0}

(vi) From the figure (vi)

Construct a line parallel to the given parallel lines

âˆ 1 = 4x and âˆ 2 = 6x are corresponding angles

âˆ 1 + âˆ 2 = 130^{0}

Substituting the values

4x + 6x = 130^{0}

10x = 130^{0}

So we get

x = 130/10 = 13^{0}